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A variant of the large sieve inequality with explicit constants - - PowerPoint PPT Presentation

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A variant of the large sieve inequality with explicit constants Maciej Grzekowiak Adam Mickiewicz University Pozna, Poland Number Theoretic Methods in Cryptology Paris 2019 MG (UAM Pozna) Sieve NutMic 2019 1 / 25 Outline 1 The large

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A variant of the large sieve inequality with explicit constants

Maciej Grześkowiak

Adam Mickiewicz University Poznań, Poland

Number Theoretic Methods in Cryptology Paris 2019

MG (UAM Poznań) Sieve NutMic 2019 1 / 25

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Outline

1 The large sieve inequality MG (UAM Poznań) Sieve NutMic 2019 2 / 25

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Outline

1 The large sieve inequality 2 The algorithmic number theory problem MG (UAM Poznań) Sieve NutMic 2019 2 / 25

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Outline

1 The large sieve inequality 2 The algorithmic number theory problem 3 Application of the large sieve inequality MG (UAM Poznań) Sieve NutMic 2019 2 / 25

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Outline

1 The large sieve inequality 2 The algorithmic number theory problem 3 Application of the large sieve inequality MG (UAM Poznań) Sieve NutMic 2019 2 / 25

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The large sieve inequality

We define S(x) =

M+N

  • n=M+1

cne(nx), e(θ) = e2πiθ, where the cn are arbitrary complex numbers.

MG (UAM Poznań) Sieve NutMic 2019 3 / 25

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The large sieve inequality

We define S(x) =

M+N

  • n=M+1

cne(nx), e(θ) = e2πiθ, where the cn are arbitrary complex numbers. The distance to nearest integer function θ = min{|θ − n| : n ∈ Z}

MG (UAM Poznań) Sieve NutMic 2019 3 / 25

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The large sieve inequality

Let x1, . . . xR be points which are well spaced modulo 1 in the sense that xr − xs ≥ δ (1) for s = r, where 0 < δ ≤ 1

2.

MG (UAM Poznań) Sieve NutMic 2019 4 / 25

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The large sieve inequality

Let x1, . . . xR be points which are well spaced modulo 1 in the sense that xr − xs ≥ δ (1) for s = r, where 0 < δ ≤ 1

2.

The large sieve is an inequality of the form

R

  • r=1

|S(xr)|2 ≤ ∆

M+N

  • n=M+1

|cn|2, (2) where ∆ = ∆(N, δ).

MG (UAM Poznań) Sieve NutMic 2019 4 / 25

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The large sieve inequality

Let x1, . . . xR be points which are well spaced modulo 1 in the sense that xr − xs ≥ δ (1) for s = r, where 0 < δ ≤ 1

2.

The large sieve is an inequality of the form

R

  • r=1

|S(xr)|2 ≤ ∆

M+N

  • n=M+1

|cn|2, (2) where ∆ = ∆(N, δ). [Gallagher] For example, we can take ‘∆ = πN + δ−1

MG (UAM Poznań) Sieve NutMic 2019 4 / 25

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Application of the large sieve inequality

Let xr = a

q be points, where (a, q) = 1 , q ≤ Q, .

If a

q = a′ q′ then

MG (UAM Poznań) Sieve NutMic 2019 5 / 25

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Application of the large sieve inequality

Let xr = a

q be points, where (a, q) = 1 , q ≤ Q, .

If a

q = a′ q′ then

  • a

q − a′ q′

  • =
  • aq′ − a′q

qq′

  • ≥

1 qq′ ≥ 1 Q2 We may take δ = Q−2, we obtain

MG (UAM Poznań) Sieve NutMic 2019 5 / 25

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Application of the large sieve inequality

Lemma

  • q≤Q

q

  • a=1

(a,q)=1

|S(a/q)|2 ≤ (N + Q2)

M+N

  • n=M+1

|cn|2, where the summation is over primes q.

MG (UAM Poznań) Sieve NutMic 2019 6 / 25

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Application of the large sieve inequality

Let π(x; a, q) = ♯{p ≤ x : p ≡ a (mod q), (a, q) = 1}

MG (UAM Poznań) Sieve NutMic 2019 7 / 25

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Application of the large sieve inequality

Let π(x; a, q) = ♯{p ≤ x : p ≡ a (mod q), (a, q) = 1} Then π(x + y; a, q) − π(x; a, q) ≤ 2y ϕ(q) log(y/q)

  • 1 + O

log log(3y/q) log(2y/q)

  • for y > q.

MG (UAM Poznań) Sieve NutMic 2019 7 / 25

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Application of the large sieve inequality

Let T(χ) =

M+N

  • n=M+1

cnχ(n) where χ is a Dirichlet character (mod q).

MG (UAM Poznań) Sieve NutMic 2019 8 / 25

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Application of the large sieve inequality

Let T(χ) =

M+N

  • n=M+1

cnχ(n) where χ is a Dirichlet character (mod q). Gallagher show ∗

χ mod q

|T(χ)|2 ≤ ϕ(q) q

q

  • a=1

(a,q)=1

|S(a/q)|2 where ∗ denotes summation over primitive multiplicative characters χ (mod q).

MG (UAM Poznań) Sieve NutMic 2019 8 / 25

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Application of the large sieve inequality

We obtain

  • q≤Q

q ϕ(q) ∗

χ mod q

|T(χ)|2 ≤ (N + Q2)

M+N

  • n=M+1

|cn|2, where the summation is over primes q and ∗ denotes summation over primitive multiplicative characters χ (mod q).

MG (UAM Poznań) Sieve NutMic 2019 9 / 25

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Generalization of the large sieve inequality

Huxley generalized to algebraic number fields K, [K : Q] = k.

MG (UAM Poznań) Sieve NutMic 2019 10 / 25

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Generalization of the large sieve inequality

Huxley generalized to algebraic number fields K, [K : Q] = k. He considered algebraic integers of α ∈ K such that α = n1ω1 + . . . + nkωk, Mi + 1 ≤ ni ≤ Mi + Ni, i = 1, . . . k,

MG (UAM Poznań) Sieve NutMic 2019 10 / 25

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Generalization of the large sieve inequality

Huxley generalized to algebraic number fields K, [K : Q] = k. He considered algebraic integers of α ∈ K such that α = n1ω1 + . . . + nkωk, Mi + 1 ≤ ni ≤ Mi + Ni, i = 1, . . . k, Schaal considered α ∈ K lying in the domains which not necessarily depend on special integer basis of K.

MG (UAM Poznań) Sieve NutMic 2019 10 / 25

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Generalization of the large sieve inequality

Huxley generalized to algebraic number fields K, [K : Q] = k. He considered algebraic integers of α ∈ K such that α = n1ω1 + . . . + nkωk, Mi + 1 ≤ ni ≤ Mi + Ni, i = 1, . . . k, Schaal considered α ∈ K lying in the domains which not necessarily depend on special integer basis of K. Hinz proved a variant of the large sieve inequality to algebraic number K

MG (UAM Poznań) Sieve NutMic 2019 10 / 25

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Problem

Find two primes p and q such that q | ♯E(Fp).

MG (UAM Poznań) Sieve NutMic 2019 11 / 25

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Problem

Find two primes p and q such that q | ♯E(Fp). Our assumptions p should be as close to q as possible

MG (UAM Poznań) Sieve NutMic 2019 11 / 25

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Problem

Find two primes p and q such that q | ♯E(Fp). Our assumptions p should be as close to q as possible works in a polynomial time with respect to p,

MG (UAM Poznań) Sieve NutMic 2019 11 / 25

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Problem

Find two primes p and q such that q | ♯E(Fp). Our assumptions p should be as close to q as possible works in a polynomial time with respect to p, give a proof without assumptions of any hypotheses, any heuristics,

MG (UAM Poznań) Sieve NutMic 2019 11 / 25

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Problem

Find two primes p and q such that q | ♯E(Fp). Our assumptions p should be as close to q as possible works in a polynomial time with respect to p, give a proof without assumptions of any hypotheses, any heuristics, compute the order of magnitude of p, q for which we can proof that the algorithm works

MG (UAM Poznań) Sieve NutMic 2019 11 / 25

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  • Application. Elliptic Curve Cryptography (ECC)

Theorem [Shparlinski, Sutherland 2014] Given a real number x > 3.There is an Algorithm that outputs p ∈ [x, 2x], a, b ∈ Fp, N = ♯E(Fp), where p is uniformly distributed over primes in [x, 2x] and the pair (a, b) is then uniformly distributed over pairs in Fp × Fp for which ♯E(Fp) is prime. Assuming the GRH, the expected running time of the Algorithm is O((log x)5(log log x)3 log log log x)

MG (UAM Poznań) Sieve NutMic 2019 12 / 25

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  • Application. Elliptic Curve Cryptography (ECC)

Theorem [Shparlinski, Sutherland 2017] Assume the GRH. There is a deterministic algorithm that, given a prime p and an integer m = o(p1/2(log p)−4), outputs an elliptic curve E(Fp) with m | ♯E(Fp) in O(mp1/2) time.

MG (UAM Poznań) Sieve NutMic 2019 13 / 25

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  • Application. Elliptic Curve Cryptography (ECC)

CM method:

MG (UAM Poznań) Sieve NutMic 2019 14 / 25

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  • Application. Elliptic Curve Cryptography (ECC)

CM method: select p,

MG (UAM Poznań) Sieve NutMic 2019 14 / 25

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  • Application. Elliptic Curve Cryptography (ECC)

CM method: select p, find ∆ < 0 and s, t ∈ Z such that 4p = t2 − ∆s2,

MG (UAM Poznań) Sieve NutMic 2019 14 / 25

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  • Application. Elliptic Curve Cryptography (ECC)

CM method: select p, find ∆ < 0 and s, t ∈ Z such that 4p = t2 − ∆s2, If p + 1 ± t is a prime, then construct E, or

MG (UAM Poznań) Sieve NutMic 2019 14 / 25

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  • Application. Elliptic Curve Cryptography (ECC)

CM method: select p, find ∆ < 0 and s, t ∈ Z such that 4p = t2 − ∆s2, If p + 1 ± t is a prime, then construct E, or If p + 1 ± t has a big prime factor q, then construct E,

MG (UAM Poznań) Sieve NutMic 2019 14 / 25

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CM-primes

MG (UAM Poznań) Sieve NutMic 2019 15 / 25

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CM-primes

DEFINITION:

MG (UAM Poznań) Sieve NutMic 2019 15 / 25

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CM-primes

DEFINITION: Primes p and q are CM-primes with respect to ∆ < 0 if

MG (UAM Poznań) Sieve NutMic 2019 15 / 25

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CM-primes

DEFINITION: Primes p and q are CM-primes with respect to ∆ < 0 if there exist integers s and t such that |t| ≤ 2√p, q|p + 1 − t, 4p − t2 = ∆s2.

MG (UAM Poznań) Sieve NutMic 2019 15 / 25

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CM-primes

DEFINITION: Primes p and q are CM-primes with respect to ∆ < 0 if there exist integers s and t such that |t| ≤ 2√p, q|p + 1 − t, 4p − t2 = ∆s2. Let p and q be CM-primes with respect to ∆.

MG (UAM Poznań) Sieve NutMic 2019 15 / 25

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CM-primes

DEFINITION: Primes p and q are CM-primes with respect to ∆ < 0 if there exist integers s and t such that |t| ≤ 2√p, q|p + 1 − t, 4p − t2 = ∆s2. Let p and q be CM-primes with respect to ∆. There exist E(Fp) such that q | ♯E(Fp) = p + 1 − t

MG (UAM Poznań) Sieve NutMic 2019 15 / 25

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CM-primes

DEFINITION: Primes p and q are CM-primes with respect to ∆ < 0 if there exist integers s and t such that |t| ≤ 2√p, q|p + 1 − t, 4p − t2 = ∆s2. Let p and q be CM-primes with respect to ∆. There exist E(Fp) such that q | ♯E(Fp) = p + 1 − t To construct E we use CM method, O(|∆|1+ǫ),

MG (UAM Poznań) Sieve NutMic 2019 15 / 25

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CM-primes

DEFINITION: Primes p and q are CM-primes with respect to ∆ < 0 if there exist integers s and t such that |t| ≤ 2√p, q|p + 1 − t, 4p − t2 = ∆s2. Let p and q be CM-primes with respect to ∆. There exist E(Fp) such that q | ♯E(Fp) = p + 1 − t To construct E we use CM method, O(|∆|1+ǫ), ∆ ≤ 1012

MG (UAM Poznań) Sieve NutMic 2019 15 / 25

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CM-primes

K = Q( √ ∆), ∆ ≡ 1 (mod 4), ω = 1+

√ ∆ 2

MG (UAM Poznań) Sieve NutMic 2019 16 / 25

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CM-primes

K = Q( √ ∆), ∆ ≡ 1 (mod 4), ω = 1+

√ ∆ 2

OK = {a + bω : a, b ∈ Z},

MG (UAM Poznań) Sieve NutMic 2019 16 / 25

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CM-primes

K = Q( √ ∆), ∆ ≡ 1 (mod 4), ω = 1+

√ ∆ 2

OK = {a + bω : a, b ∈ Z}, Input: n, m ∈ N, (m, n) = 1, x ∈ R,

MG (UAM Poznań) Sieve NutMic 2019 16 / 25

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CM-primes

K = Q( √ ∆), ∆ ≡ 1 (mod 4), ω = 1+

√ ∆ 2

OK = {a + bω : a, b ∈ Z}, Input: n, m ∈ N, (m, n) = 1, x ∈ R, γ = f + gω ∈ OK, |f |, |g| ≤ n, NK/Q(γ) ≡ m (mod n)

MG (UAM Poznań) Sieve NutMic 2019 16 / 25

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CM-primes

K = Q( √ ∆), ∆ ≡ 1 (mod 4), ω = 1+

√ ∆ 2

OK = {a + bω : a, b ∈ Z}, Input: n, m ∈ N, (m, n) = 1, x ∈ R, γ = f + gω ∈ OK, |f |, |g| ≤ n, NK/Q(γ) ≡ m (mod n) Procedure FindPrimeQ (MG)

MG (UAM Poznań) Sieve NutMic 2019 16 / 25

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CM-primes

K = Q( √ ∆), ∆ ≡ 1 (mod 4), ω = 1+

√ ∆ 2

OK = {a + bω : a, b ∈ Z}, Input: n, m ∈ N, (m, n) = 1, x ∈ R, γ = f + gω ∈ OK, |f |, |g| ≤ n, NK/Q(γ) ≡ m (mod n) Procedure FindPrimeQ (MG)

1 Find

|u| ≤ (

√ 1−∆ √ −4∆(2x)1/2 − f )n−1,

|v| ≤ (

1 √ −∆(2x)1/2 − g)n−1

MG (UAM Poznań) Sieve NutMic 2019 16 / 25

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SLIDE 49

CM-primes

K = Q( √ ∆), ∆ ≡ 1 (mod 4), ω = 1+

√ ∆ 2

OK = {a + bω : a, b ∈ Z}, Input: n, m ∈ N, (m, n) = 1, x ∈ R, γ = f + gω ∈ OK, |f |, |g| ≤ n, NK/Q(γ) ≡ m (mod n) Procedure FindPrimeQ (MG)

1 Find

|u| ≤ (

√ 1−∆ √ −4∆(2x)1/2 − f )n−1,

|v| ≤ (

1 √ −∆(2x)1/2 − g)n−1

2 Compute α = nu + f + (nv + g)ω MG (UAM Poznań) Sieve NutMic 2019 16 / 25

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SLIDE 50

CM-primes

K = Q( √ ∆), ∆ ≡ 1 (mod 4), ω = 1+

√ ∆ 2

OK = {a + bω : a, b ∈ Z}, Input: n, m ∈ N, (m, n) = 1, x ∈ R, γ = f + gω ∈ OK, |f |, |g| ≤ n, NK/Q(γ) ≡ m (mod n) Procedure FindPrimeQ (MG)

1 Find

|u| ≤ (

√ 1−∆ √ −4∆(2x)1/2 − f )n−1,

|v| ≤ (

1 √ −∆(2x)1/2 − g)n−1

2 Compute α = nu + f + (nv + g)ω 3 If q = NK/Q(α) is a prime, MG (UAM Poznań) Sieve NutMic 2019 16 / 25

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CM-primes

K = Q( √ ∆), ∆ ≡ 1 (mod 4), ω = 1+

√ ∆ 2

OK = {a + bω : a, b ∈ Z}, Input: n, m ∈ N, (m, n) = 1, x ∈ R, γ = f + gω ∈ OK, |f |, |g| ≤ n, NK/Q(γ) ≡ m (mod n) Procedure FindPrimeQ (MG)

1 Find

|u| ≤ (

√ 1−∆ √ −4∆(2x)1/2 − f )n−1,

|v| ≤ (

1 √ −∆(2x)1/2 − g)n−1

2 Compute α = nu + f + (nv + g)ω 3 If q = NK/Q(α) is a prime, then RETURN α = a + bω and q MG (UAM Poznań) Sieve NutMic 2019 16 / 25

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CM-primes

Input: K = Q( √ ∆), α = a + bω ∈ OK, q = NK/Q(α) ≡ m (mod n), ∆ ≡ 1 (mod 4), 0 < ε < 2/5

MG (UAM Poznań) Sieve NutMic 2019 17 / 25

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CM-primes

Input: K = Q( √ ∆), α = a + bω ∈ OK, q = NK/Q(α) ≡ m (mod n), ∆ ≡ 1 (mod 4), 0 < ε < 2/5 Procedure FindPrimeP (MG)

MG (UAM Poznań) Sieve NutMic 2019 17 / 25

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CM-primes

Input: K = Q( √ ∆), α = a + bω ∈ OK, q = NK/Q(α) ≡ m (mod n), ∆ ≡ 1 (mod 4), 0 < ε < 2/5 Procedure FindPrimeP (MG)

1 Find

|s| ≤ √ 1 − ∆ √ −4∆ (2x)(3+5ε)/(4−10ε), |t| ≤ 1 √ −∆(2x)(3+5ε)/(4−10ε).

MG (UAM Poznań) Sieve NutMic 2019 17 / 25

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CM-primes

Input: K = Q( √ ∆), α = a + bω ∈ OK, q = NK/Q(α) ≡ m (mod n), ∆ ≡ 1 (mod 4), 0 < ε < 2/5 Procedure FindPrimeP (MG)

1 Find

|s| ≤ √ 1 − ∆ √ −4∆ (2x)(3+5ε)/(4−10ε), |t| ≤ 1 √ −∆(2x)(3+5ε)/(4−10ε).

2 Compute β = as − 1−∆

4 bt + 1 + (bs + (a + b)t)ω

MG (UAM Poznań) Sieve NutMic 2019 17 / 25

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CM-primes

Input: K = Q( √ ∆), α = a + bω ∈ OK, q = NK/Q(α) ≡ m (mod n), ∆ ≡ 1 (mod 4), 0 < ε < 2/5 Procedure FindPrimeP (MG)

1 Find

|s| ≤ √ 1 − ∆ √ −4∆ (2x)(3+5ε)/(4−10ε), |t| ≤ 1 √ −∆(2x)(3+5ε)/(4−10ε).

2 Compute β = as − 1−∆

4 bt + 1 + (bs + (a + b)t)ω

3 Compute p = NK/Q(β) MG (UAM Poznań) Sieve NutMic 2019 17 / 25

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SLIDE 57

CM-primes

Input: K = Q( √ ∆), α = a + bω ∈ OK, q = NK/Q(α) ≡ m (mod n), ∆ ≡ 1 (mod 4), 0 < ε < 2/5 Procedure FindPrimeP (MG)

1 Find

|s| ≤ √ 1 − ∆ √ −4∆ (2x)(3+5ε)/(4−10ε), |t| ≤ 1 √ −∆(2x)(3+5ε)/(4−10ε).

2 Compute β = as − 1−∆

4 bt + 1 + (bs + (a + b)t)ω

3 Compute p = NK/Q(β) 4 If p < x or p > (2x)5/(2−5ε) is a prime, then MG (UAM Poznań) Sieve NutMic 2019 17 / 25

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CM-primes

Input: K = Q( √ ∆), α = a + bω ∈ OK, q = NK/Q(α) ≡ m (mod n), ∆ ≡ 1 (mod 4), 0 < ε < 2/5 Procedure FindPrimeP (MG)

1 Find

|s| ≤ √ 1 − ∆ √ −4∆ (2x)(3+5ε)/(4−10ε), |t| ≤ 1 √ −∆(2x)(3+5ε)/(4−10ε).

2 Compute β = as − 1−∆

4 bt + 1 + (bs + (a + b)t)ω

3 Compute p = NK/Q(β) 4 If p < x or p > (2x)5/(2−5ε) is a prime, then 5 RETURN β = c + dω, p. MG (UAM Poznań) Sieve NutMic 2019 17 / 25

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SLIDE 59

CM-primes

Input: K = Q( √ ∆), α = a + bω ∈ OK, q = NK/Q(α) ≡ m (mod n), ∆ ≡ 1 (mod 4), 0 < ε < 2/5 Procedure FindPrimeP (MG)

1 Find

|s| ≤ √ 1 − ∆ √ −4∆ (2x)(3+5ε)/(4−10ε), |t| ≤ 1 √ −∆(2x)(3+5ε)/(4−10ε).

2 Compute β = as − 1−∆

4 bt + 1 + (bs + (a + b)t)ω

3 Compute p = NK/Q(β) 4 If p < x or p > (2x)5/(2−5ε) is a prime, then 5 RETURN β = c + dω, p.

β ≡ 1 (mod α),

MG (UAM Poznań) Sieve NutMic 2019 17 / 25

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SLIDE 60

CM-primes

Input: K = Q( √ ∆), α = a + bω ∈ OK, q = NK/Q(α) ≡ m (mod n), ∆ ≡ 1 (mod 4), 0 < ε < 2/5 Procedure FindPrimeP (MG)

1 Find

|s| ≤ √ 1 − ∆ √ −4∆ (2x)(3+5ε)/(4−10ε), |t| ≤ 1 √ −∆(2x)(3+5ε)/(4−10ε).

2 Compute β = as − 1−∆

4 bt + 1 + (bs + (a + b)t)ω

3 Compute p = NK/Q(β) 4 If p < x or p > (2x)5/(2−5ε) is a prime, then 5 RETURN β = c + dω, p.

β ≡ 1 (mod α), NK/Q(α) | NK/Q(β − 1) = NK/Q(β) + 1 − Tr(β)

MG (UAM Poznań) Sieve NutMic 2019 17 / 25

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SLIDE 61

CM-primes

Theorem [M.G.2015] There exists x0 > 0 such that for every x ≥ x0 and λ ∈ R, λ ≥ 1,

MG (UAM Poznań) Sieve NutMic 2019 18 / 25

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SLIDE 62

CM-primes

Theorem [M.G.2015] There exists x0 > 0 such that for every x ≥ x0 and λ ∈ R, λ ≥ 1, procedure FindPrimeQ finds α ∈ OK, q = NK/Q(α) ≡ m (mod n), x ≤ NK/Q(α) ≤ 2x

MG (UAM Poznań) Sieve NutMic 2019 18 / 25

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SLIDE 63

CM-primes

Theorem [M.G.2015] There exists x0 > 0 such that for every x ≥ x0 and λ ∈ R, λ ≥ 1, procedure FindPrimeQ finds α ∈ OK, q = NK/Q(α) ≡ m (mod n), x ≤ NK/Q(α) ≤ 2x with probability greater than or equal to 1 − e−λ

MG (UAM Poznań) Sieve NutMic 2019 18 / 25

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SLIDE 64

CM-primes

Theorem [M.G.2015] There exists x0 > 0 such that for every x ≥ x0 and λ ∈ R, λ ≥ 1, procedure FindPrimeQ finds α ∈ OK, q = NK/Q(α) ≡ m (mod n), x ≤ NK/Q(α) ≤ 2x with probability greater than or equal to 1 − e−λ after repeating [c1λ(log x)] steps,

MG (UAM Poznań) Sieve NutMic 2019 18 / 25

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SLIDE 65

CM-primes

Theorem [M.G.2015] There exists x0 > 0 such that for every x ≥ x0 and λ ∈ R, λ ≥ 1, procedure FindPrimeQ finds α ∈ OK, q = NK/Q(α) ≡ m (mod n), x ≤ NK/Q(α) ≤ 2x with probability greater than or equal to 1 − e−λ after repeating [c1λ(log x)] steps, where c1 =

4 √ 1−∆h∗

f (K)

−∆n2

, f = nOK.

MG (UAM Poznań) Sieve NutMic 2019 18 / 25

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SLIDE 66

CM-primes

Theorem [M.G.2015] There exists x0 > 0 such that for every x ≥ x0 and λ ∈ R, λ ≥ 1, procedure FindPrimeQ finds α ∈ OK, q = NK/Q(α) ≡ m (mod n), x ≤ NK/Q(α) ≤ 2x with probability greater than or equal to 1 − e−λ after repeating [c1λ(log x)] steps, where c1 =

4 √ 1−∆h∗

f (K)

−∆n2

, f = nOK.

MG (UAM Poznań) Sieve NutMic 2019 18 / 25

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SLIDE 67

CM-primes

Theorem [MG 2015] Let α ∈ OK, x ≤ q = NK/Q(α) ≤ 2x.

MG (UAM Poznań) Sieve NutMic 2019 19 / 25

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SLIDE 68

CM-primes

Theorem [MG 2015] Let α ∈ OK, x ≤ q = NK/Q(α) ≤ 2x. there exists x0 > 0 such that for every x ≥ x0, and for λ ∈ R, λ ≥ 1, and for any A > 2

MG (UAM Poznań) Sieve NutMic 2019 19 / 25

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SLIDE 69

CM-primes

Theorem [MG 2015] Let α ∈ OK, x ≤ q = NK/Q(α) ≤ 2x. there exists x0 > 0 such that for every x ≥ x0, and for λ ∈ R, λ ≥ 1, and for any A > 2 procedure FindPrimeP finds β = c + dω, p = NK/Q(β), x ≤ NK/Q(β) ≤ (2x)5/(2−5ε),

MG (UAM Poznań) Sieve NutMic 2019 19 / 25

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SLIDE 70

CM-primes

Theorem [MG 2015] Let α ∈ OK, x ≤ q = NK/Q(α) ≤ 2x. there exists x0 > 0 such that for every x ≥ x0, and for λ ∈ R, λ ≥ 1, and for any A > 2 procedure FindPrimeP finds β = c + dω, p = NK/Q(β), x ≤ NK/Q(β) ≤ (2x)5/(2−5ε), with probability greater than or equal to 1 − e−λ after repeating [c2λ(log 2x)] steps of the procedure,

MG (UAM Poznań) Sieve NutMic 2019 19 / 25

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SLIDE 71

CM-primes

Theorem [MG 2015] Let α ∈ OK, x ≤ q = NK/Q(α) ≤ 2x. there exists x0 > 0 such that for every x ≥ x0, and for λ ∈ R, λ ≥ 1, and for any A > 2 procedure FindPrimeP finds β = c + dω, p = NK/Q(β), x ≤ NK/Q(β) ≤ (2x)5/(2−5ε), with probability greater than or equal to 1 − e−λ after repeating [c2λ(log 2x)] steps of the procedure, for almost all α with the possible exception of at most O(x(log x)−A) values of α, where c2 =

40h(K) √ 1−∆ −(2−5ε)w(K)∆

MG (UAM Poznań) Sieve NutMic 2019 19 / 25

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SLIDE 72

Notation

K - any totally imaginary algebraic number field of degree [K : Q] = 2r2,

MG (UAM Poznań) Sieve NutMic 2019 20 / 25

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SLIDE 73

Notation

K - any totally imaginary algebraic number field of degree [K : Q] = 2r2, f - a given non-zero integral ideal of the ring of OK,

MG (UAM Poznań) Sieve NutMic 2019 20 / 25

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SLIDE 74

Notation

K - any totally imaginary algebraic number field of degree [K : Q] = 2r2, f - a given non-zero integral ideal of the ring of OK, H (mod f) - any ideal class mod f in the ”narrow” sense,

MG (UAM Poznań) Sieve NutMic 2019 20 / 25

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SLIDE 75

Notation

K - any totally imaginary algebraic number field of degree [K : Q] = 2r2, f - a given non-zero integral ideal of the ring of OK, H (mod f) - any ideal class mod f in the ”narrow” sense, h∗

f (K) - the number of elements of H,

MG (UAM Poznań) Sieve NutMic 2019 20 / 25

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SLIDE 76

Notation

K - any totally imaginary algebraic number field of degree [K : Q] = 2r2, f - a given non-zero integral ideal of the ring of OK, H (mod f) - any ideal class mod f in the ”narrow” sense, h∗

f (K) - the number of elements of H,

Let s = σ + it ζ(s, χ) =

  • a∈OK

χ(a) (Na)s , σ > 1, where a runs through integral ideals of OK

MG (UAM Poznań) Sieve NutMic 2019 20 / 25

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SLIDE 77

Notation

χ0 denote the principal character modulo f

MG (UAM Poznań) Sieve NutMic 2019 21 / 25

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SLIDE 78

Notation

χ0 denote the principal character modulo f Let E0 = E0(χ) = 1 for χ = χ0 for χ = χ0

MG (UAM Poznań) Sieve NutMic 2019 21 / 25

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SLIDE 79

Notation

χ0 denote the principal character modulo f Let E0 = E0(χ) = 1 for χ = χ0 for χ = χ0 Fix X mod f ∈ H.

MG (UAM Poznań) Sieve NutMic 2019 21 / 25

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SLIDE 80

Notation

χ0 denote the principal character modulo f Let E0 = E0(χ) = 1 for χ = χ0 for χ = χ0 Fix X mod f ∈ H. Ψ(x, X) =

  • x≤Npm≤2x

pm∈X

log Np, where p runs through prime ideals of OK

MG (UAM Poznań) Sieve NutMic 2019 21 / 25

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SLIDE 81

Theorem [M. G. 2017]

If |∆| ≥ 9 and there is no zero in the region σ ≥ 1 − 0.0795

  • log |∆| + 0.7761 log
  • (|t| + 1)2r2(Nf)1−E0(χ)−1

, then

MG (UAM Poznań) Sieve NutMic 2019 22 / 25

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SLIDE 82

Theorem [M. G. 2017]

If |∆| ≥ 9 and there is no zero in the region σ ≥ 1 − 0.0795

  • log |∆| + 0.7761 log
  • (|t| + 1)2r2(Nf)1−E0(χ)−1

, then Ψ(x, X) ≥ x 2h∗

f (K),

MG (UAM Poznań) Sieve NutMic 2019 22 / 25

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SLIDE 83

Theorem [M. G. 2017]

If |∆| ≥ 9 and there is no zero in the region σ ≥ 1 − 0.0795

  • log |∆| + 0.7761 log
  • (|t| + 1)2r2(Nf)1−E0(χ)−1

, then Ψ(x, X) ≥ x 2h∗

f (K),

for log x ≥

  • 23.441√r2
  • 1 + (2 log (17.252C√r2))

1 2 + 2

3 log (17.252C√r2) 2 ,

MG (UAM Poznań) Sieve NutMic 2019 22 / 25

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SLIDE 84

Theorem [M. G. 2017]

If |∆| ≥ 9 and there is no zero in the region σ ≥ 1 − 0.0795

  • log |∆| + 0.7761 log
  • (|t| + 1)2r2(Nf)1−E0(χ)−1

, then Ψ(x, X) ≥ x 2h∗

f (K),

for log x ≥

  • 23.441√r2
  • 1 + (2 log (17.252C√r2))

1 2 + 2

3 log (17.252C√r2) 2 , where C = (3056|∆|

1.933 r2

+ 15382.485|∆|

1.289 r2 (Nf) 1 r2 h∗

f (K))r2 2 log(|∆|Nf).

MG (UAM Poznań) Sieve NutMic 2019 22 / 25

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SLIDE 85

Notation

Let D < 0 be a square-free integer, Let K = Q( √ D) with OK = {a + bω : a, b ∈ Z}, where ω =

  • 1+

√ D 2

if D ≡ 1 (mod 4) √ D if D ≡ 2, 3 (mod 4) Let x > 1 , we define R = {β ∈ OK : |β| < √x}.

MG (UAM Poznań) Sieve NutMic 2019 23 / 25

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SLIDE 86

Theorem [M. G. 2019]

Fix Q > 1,

  • Nq≤Q

(q,a)=1

′

σ mod q

  • α∈R

α≡0 (mod a)

c(α)σ(α)

  • 2

≤ f (x, a, Q)

  • α∈R

α≡0 (mod a)

|c(α)|2,

MG (UAM Poznań) Sieve NutMic 2019 24 / 25

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SLIDE 87

Theorem [M. G. 2019]

Fix Q > 1,

  • Nq≤Q

(q,a)=1

′

σ mod q

  • α∈R

α≡0 (mod a)

c(α)σ(α)

  • 2

≤ f (x, a, Q)

  • α∈R

α≡0 (mod a)

|c(α)|2, where f (x, a, Q) = √

8

4

√ 3

x

Na

1

4 + c0|D| 1 4 Q 1 2

4 , c0 =     

  • 1 −

1 √ 3

− 1

2

if D ≡ 1 (mod 4),

  • 1

2 − 1 2 √ 3

− 1

2

if D ≡ 2, 3 (mod 4), and ′ denotes summation over primitive additive characters (mod q), and the c(α) are any complex number.

MG (UAM Poznań) Sieve NutMic 2019 24 / 25

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SLIDE 88

Theorem [M. G. 2019]

Fix Q > 1. We have

  • Nq≤Q

Nq Φ(q) ∗

χ mod q

  • α∈R

α≡0 (mod a)

c(α)χ(α)

  • 2

≤ f (x, a, Q)

  • α∈R

α≡0 (mod a)

|c(α)|2,

MG (UAM Poznań) Sieve NutMic 2019 25 / 25

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SLIDE 89

Theorem [M. G. 2019]

Fix Q > 1. We have

  • Nq≤Q

Nq Φ(q) ∗

χ mod q

  • α∈R

α≡0 (mod a)

c(α)χ(α)

  • 2

≤ f (x, a, Q)

  • α∈R

α≡0 (mod a)

|c(α)|2, where f (x, a, Q) = √

8

4

√ 3

x

Na

1

4 + c0|D| 1 4 Q 1 2

4 , c0 =     

  • 1 −

1 √ 3

− 1

2

if D ≡ 1 (mod 4),

  • 1

2 − 1 2 √ 3

− 1

2

if D ≡ 2, 3 (mod 4), and ∗ denotes summation over primitive multiplicative characters (mod q), and the c(α) are any complex numbers.

MG (UAM Poznań) Sieve NutMic 2019 25 / 25

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SLIDE 90

Thank you

MG (UAM Poznań) Sieve NutMic 2019 26 / 25